Give reasons for the following:

(a) Transition metals show variable oxidation states.

(b) Evalue for (Zn²⁺/Zn) is negative while that of (Cu²⁺/Cu) is positive.

(c) Higher oxidation state of Mn with fluorine is +4 whereas with oxygen is +7.

(a) Transition metals show variable oxidation states because of small energy difference between (n-1)d and ns orbitals and as a result both (n-1)d and ns electrons take part in bond formation.

(b) The electrode potential of metals is decided by the following factors:

1. Enthalpy of atomisation

2. Ionisation enthalpy

3. Hydration enthalpy

The unique behaviour of Cu, having a positive E° is due to its high energy of atomization and low hydration enthalpy. The high energy to transform Cu(s) to Cu²⁺ (aq) is not balanced by its hydration enthalpy. Therefore, Evalue for (Cu²⁺/Cu) is positive.

(c) Higher oxidation state of Mn with fluorine is +4 whereas with oxygen is +7. Both O and F due to their small size and highest electronegativity can oxidize the transition metal to their highest oxidation state as it can easily unpair the electrons of metal. Among O and F, O is more superior to stabilize the highest oxidation state (+7 for Mn) of metal due to their ability to form multiple bonds.